This document contains 26 math problems involving derivatives of logarithmic, exponential, and inverse trigonometric functions. The problems include finding derivatives of expressions, setting up and solving differential equations, and determining relationships between derivatives.
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http://sandymillin.wordpress.com/iateflwebinar2024
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1. January 27, 2005 11:44 L24-ch04 Sheet number 1 Page number 127 black
CHAPTER 4
Derivatives of Logarithmic, Exponential, and
Inverse Trigonometric Functions
EXERCISE SET 4.1
2
1. y = (2x − 5)1/3 ; dy/dx = (2x − 5)−2/3
3
1 −2/3 2 −2/3
2. dy/dx = 2 + tan(x2 ) sec2 (x2 )(2x) = x sec2 (x2 ) 2 + tan(x2 )
3 3
−1/3
2 x+1 x − 2 − (x + 1) 2
3. dy/dx = =−
3 x−2 (x − 2)2 (x + 1)1/3 (x − 2)5/3
−1/2 −1/2
1 x2 + 1 d x2 + 1 1 x2 + 1 −12x 6x
4. dy/dx = = =− √
2 x2 − 5 dx x2−5 2 x2 − 5 (x2 − 5)2 (x2 − 5)3/2 x2 + 1
2 1 2
5. dy/dx = x3 − (5x2 + 1)−5/3 (10x) + 3x2 (5x2 + 1)−2/3 = x (5x2 + 1)−5/3 (25x2 + 9)
3 3
√
3
2x − 1 1 2 −4x + 3
6. dy/dx = − + = 2
x2 x 3(2x − 1)2/3 3x (2x − 1)2/3
5 15[sin(3/x)]3/2 cos(3/x)
7. dy/dx = [sin(3/x)]3/2 [cos(3/x)](−3/x2 ) = −
2 2x2
1 −3/2 3 2 −3/2
8. dy/dx = − cos(x3 ) − sin(x3 ) (3x2 ) = x sin(x3 ) cos(x3 )
2 2
dy dy 6x2 − y − 1
9. (a) 1 + y + x − 6x2 = 0, =
dx dx x
2 + 2x3 − x 2 dy 2
(b) y = = + 2x2 − 1, = − 2 + 4x
x x dx x
dy 1 1 1 1 2 2
(c) From Part (a), = 6x − − y = 6x − − + 2x2 − 1 = 4x −
dx x x x x x x2
1 −1/2 dy dy √
10. (a) y − cos x = 0 or = 2 y cos x
2 dx dx
dy
(b) y = (2 + sin x) = 4 + 4 sin x + sin2 x so
2
= 4 cos x + 2 sin x cos x
dx
dy √
(c) from Part (a), = 2 y cos x = 2 cos x(2 + sin x) = 4 cos x + 2 sin x cos x
dx
dy dy x
11. 2x + 2y = 0 so =−
dx dx y
dy dy dy 3y 2 − 3x y 2 − x2
12. 3x2 + 3y 2 = 3y 2 + 6xy , = 2 = 2
dx dx dx 3y − 6xy y − 2xy
dy dy
13. x2 + 2xy + 3x(3y 2 ) + 3y 3 − 1 = 0
dx dx
dy dy 1 − 2xy − 3y 3
(x2 + 9xy 2 ) = 1 − 2xy − 3y 3 so =
dx dx x2 + 9xy 2
127
2. January 27, 2005 11:44 L24-ch04 Sheet number 2 Page number 128 black
128 Chapter 4
dy dy
14. x3 (2y) + 3x2 y 2 − 5x2 − 10xy + 1 = 0
dx dx
dy dy 10xy − 3x2 y 2 − 1
(2x3 y − 5x2 ) = 10xy − 3x2 y 2 − 1 so =
dx dx 2x3 y − 5x2
dy
1 dy y 3/2
15. − − dx = 0, = − 3/2
2x3/2 2y 3/2 dx x
(x − y)(1 + dy/dx) − (x + y)(1 − dy/dx)
16. 2x = ,
(x − y)2
dy dy x(x − y)2 + y
2x(x − y)2 = −2y + 2x so =
dx dx x
dy dy 1 − 2xy 2 cos(x2 y 2 )
17. cos(x2 y 2 ) x2 (2y) + 2xy 2 = 1, =
dx dx 2x2 y cos(x2 y 2 )
dy dy dy y 2 sin(xy 2 )
18. − sin(xy 2 ) y 2 + 2xy = , =−
dx dx dx 2xy sin(xy 2 ) + 1
dy dy
19. 3 tan2 (xy 2 + y) sec2 (xy 2 + y) 2xy + y2 + =1
dx dx
dy 1 − 3y 2 tan2 (xy 2 + y) sec2 (xy 2 + y)
so =
dx 3(2xy + 1) tan2 (xy 2 + y) sec2 (xy 2 + y)
(1 + sec y)[3xy 2 (dy/dx) + y 3 ] − xy 3 (sec y tan y)(dy/dx) dy
20. = 4y 3 ,
(1 + sec y)2 dx
dy
multiply through by (1 + sec y)2 and solve for to get
dx
dy y(1 + sec y)
=
dx 4y(1 + sec y)2 − 3x(1 + sec y) + xy sec y tan y
2
dy dy 2x dy d2 y
21. 4x − 6y = 0, = , 4−6 − 6y = 0,
dx dx 3y dx dx2
2
d2 y 3 dy
dx −2 2(3y 2 − 2x2 ) 8
=− = =− 3
dx2 3y 9y 3 9y
dy x2 d2 y y 2 (2x) − x2 (2ydy/dx) 2xy 2 − 2x2 y(−x2 /y 2 ) 2x(y 3 + x3 )
22. = − 2, =− =− =− ,
dx y dx2 y4 y4 y5
d2 y 2x
but x3 + y 3 = 1 so =− 5
dx2 y
dy y d2 y x(dy/dx) − y(1) x(−y/x) − y 2y
23. =− , =− =− = 2
dx x dx2 x2 x2 x
2
dy dy dy y dy d2 y dy d2 y d2 y 2y(x + y)
24. y + x + 2y = 0, =− ,2 +x 2 +2 + 2y 2
= 0, =
dx dx dx x + 2y dx dx dx dx dx2 (x + 2y)3
dy d2 y dy sin y
25. = (1 + cos y)−1 , = −(1 + cos y)−2 (− sin y) =
dx dx2 dx (1 + cos y)3
3. January 27, 2005 11:44 L24-ch04 Sheet number 3 Page number 129 black
Exercise Set 4.1 129
dy cos y
26. = ,
dx 1 + x sin y
d2 y (1 + x sin y)(− sin y)(dy/dx) − (cos y)[(x cos y)(dy/dx) + sin y]
=
dx2 (1 + x sin y)2
2 sin y cos y + (x cos y)(2 sin2 y + cos2 y)
=− ,
(1 + x sin y)3
but x cos y = y, 2 sin y cos y = sin 2y, and sin2 y + cos2 y = 1 so
d2 y sin 2y + y(sin2 y + 1)
=−
dx2 (1 + x sin y)3
dy x √ dy √
27. By implicit differentiation, 2x + 2y(dy/dx) = 0, = − ; at (1/2, 3/2), = − 3/3; at
dx y dx
√ dy √ √ dy −x
(1/2, − 3/2), = + 3/3. Directly, at the upper point y = 1 − x 2, = √ =
dx dx 1 − x2
1/2 √ √ dy x √
− = −1/ 3 and at the lower point y = − 1 − x2 , =√ = +1/ 3.
3/4 dx 1−x 2
√ √
28. If y 2 − x + 1 = 0, then y = x − 1 goes through the point (10, 3) so dy/dx = 1/(2 x − √ By1).
implicit differentiation dy/dx = 1/(2y). In both cases, dy/dx|(10,3) = 1/6. Similarly y = − x − 1
√
goes through (10, −3) so dy/dx = −1/(2 x − 1) = −1/6 which yields dy/dx = 1/(2y) = −1/6.
dy dy x3 1
29. 4x3 + 4y 3 = 0, so = − 3 = − 3/4 ≈ −0.1312.
dx dx y 15
dy dy dy dy y+1
30. 3y 2 + x2 + 2xy + 2x − 6y = 0, so = −2x 2 = 0 at x = 0
dx dx dx dx 3y + x2 − 6y
dy dy
31. 4(x2 + y 2 ) 2x + 2y = 25 2x − 2y ,
dx dx
dy x[25 − 4(x2 + y 2 )] dy
= ; at (3, 1) = −9/13
dx y[25 + 4(x2 + y 2 )] dx
2 dy dy y 1/3 √ √
32. x−1/3 + y −1/3 = 0, = − 1/3 = 3 at (−1, 3 3)
3 dx dx x
da da da da 2t3 + 3a2
33. 4a3 − 4t3 = 6 a2 + 2at , solve for to get = 3
dt dt dt dt 2a − 6at
√
1 −1/2 du 1 −1/2 du u
34. u + v = 0 so = −√
2 dv 2 dv v
dω dω b2 λ dx dx 1
35. 2a2 ω + 2b2 λ = 0 so =− 2 36. 1 = (cos x) so = = sec x
dλ dλ a ω dy dy cos x
4. January 27, 2005 11:44 L24-ch04 Sheet number 4 Page number 130 black
130 Chapter 4
37. (a) y
2
x
–4 4
–2
dy dy
(b) Implicit differentiation of the equation of the curve yields (4y 3 + 2y) = 2x − 1 so =0
dx dx
only if x = 1/2 but y 4 + y 2 ≥ 0, so x = 1/2 is impossible.
1± 1 + 4y 2 + 4y 4
(c) x2 − x − (y 4 + y 2 ) = 0, so by the Quadratic Formula x = = 1 + y 2 , −y 2
2
which gives the parabolas x = 1 + y 2 , x = −y 2 .
38. (a) y
2
x
0 1 2
–2
dy dy
(b) 2y = (x − a)(x − b) + x(x − b) + x(x − a) = 3x2 − 2(a + b)x + ab. If = 0 then
dx dx
3x2 − 2(a + b)x + ab = 0. By the Quadratic Formula
2(a + b) ± 4(a + b)2 − 4 · 3ab 1
x= = a + b ± (a2 + b2 − ab)1/2 .
6 3
(c) y = ± x(x − a)(x − b). The square root is only defined for nonnegative arguments, so it is
necessary that all three of the factors x, x − a, x − b be nonnegative, or that two of them be
nonpositive. If, for example, 0 < a < b then the function is defined on the disjoint intervals
0 < x < a and b < x < +∞, so there are two parts.
39. (a) y (b) x ≈ ±1.1547
2
x
–2 2
–2
dy dy dy y − 2x dy
(c) Implicit differentiation yields 2x − x − y + 2y = 0. Solve for = . If =0
dx dx dx 2y − x dx
2
then y − 2x = 0 or y = 2x. Thus 4 = x2 − xy + y 2 = x2 − 2x2 + 4x2 = 3x2 , x = ± √ .
3
40. (a) See Exercise 39 (a)
(b) Since the equation is symmetric in x and y, we obtain, as in Exercise 39, x ≈ ±1.1547.
5. January 27, 2005 11:44 L24-ch04 Sheet number 5 Page number 131 black
Exercise Set 4.1 131
dy dy dx 2y − x dx
(c) Implicit differentiation yields 2x − x − y + 2y = 0. Solve for = . If =0
dx dx dy y − 2x dy
2 4
then 2y − x = 0 or x = 2y. Thus 4 = 4y 2 − 2y 2 + y 2 = 3y 2 , y = ± √ , x = 2y = ± √ .
3 3
41. Solve the simultaneous equations y = x, x2 −xy+y 2 = 4 to get x2 −x2 +x2 = 4, x = ±2, y = x = ±2,
so the points of intersection are (2, 2) and (−2, −2).
dy y − 2x dy dy
From Exercise 39 part (c), = . When x = y = 2, = −1; when x = y = −2, = −1;
dx 2y − x dx dx
the slopes are equal.
42. Suppose a2 − 2ab + b2 = 4. Then (−a)2 − 2(−a)(−b) + (−b)2 = a2 − 2ab + b2 = 4 so if P (a, b) lies
on C then so does Q(−a, −b).
dy y − 2x dy b − 2a
From Exercise 39 part (c), = . When x = a, y = b then = , and when
dx 2y − x dx 2b − a
dy b − 2a
x = −a, y = −b, then = , so the slopes at P and Q are equal.
dx 2b − a
43. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use
dy 2xy 2 4
implicit differentiation to get =− 2 so at (1,1), − = − , 1 + 2a = 3/2, a = 1/4
dx x + 2ay 1 + 2a 3
and hence b = 1 + 1/4 = 5/4.
44. The slope of the line x + 2y − 2 = 0 is m1 = −1/2, so the line perpendicular has slope m = 2
(negative reciprocal). The slope of the curve y 3 = 2x2 can be obtained by implicit differentiation:
dy dy 4x dy 4x
3y 2 = 4x, = 2 . Set = 2; 2 = 2, x = (3/2)y 2 . Use this in the equation of the curve:
dx dx 3y dx 3y
2
3 2 2
y 3 = 2x2 = 2((3/2)y 2 )2 = (9/2)y 4 , y = 2/9, x = = .
2 9 27
45. (a) y (b) x ≈ 0.84
2
x
–3 –1 2
–1
–3
(c) Use implicit differentiation to get dy/dx = (2y −3x2 )/(3y 2 −2x), so dy/dx = 0 if y = (3/2)x2 .
Substitute this into x3 − 2xy + y 3 = 0 to obtain 27x6 − 16x3 = 0, x3 = 16/27, x = 24/3 /3
and hence y = 25/3 /3.
46. (a) y (b) Evidently the tangent line at the point
2
x = 1, y = 1 has slope −1.
x
–3 –1 2
–1
–3
6. January 27, 2005 11:44 L24-ch04 Sheet number 6 Page number 132 black
132 Chapter 4
(c) Use implicit differentiation to get dy/dx = (2y −3x2 )/(3y 2 −2x), so dy/dx = −1 if 2y −3x2 =
−3y 2 +2x, 2(y−x)+3(y−x)(y+x) = 0. One solution is y = x; this together with x3 +y 3 = 2xy
yields x = y = 1. For these values dy/dx = −1, so that (1, 1) is a solution.
To prove that there is no other solution, suppose y = x. From dy/dx = −1 it follows
that 2(y − x) + 3(y − x)(y + x) = 0. But y = x, so x + y = −2/3. Then x3 + y 3 =
(x + y)(x2 − xy + y 2 ) = 2xy, so replacing x + y with −2/3 we get x2 + 2xy + y 2 = 0, or
(x + y)2 = 0, so y = −x. Substitute that into x3 + y 3 = 2xy to obtain x3 − x3 = −2x2 , x = 0.
But at x = y = 0 the derivative is not defined.
47. (a) The curve is the circle (x − 2)2 + y 2 = 1 about the point (2, 0) of radius 1. One tangent
line is tangent at a point P(x,y) in the first quadrant. Let Q(2, 0) be the center of the
circle. Then OP Q is a right angle, with sides |P Q| = r = 1 and |OP | = x2 + y 2 . By
the Pythagorean Theorem x2 + y 2 + 12 = 22 . Substitute this into (x − 2)2 + y 2 = 1 to
√ √
obtain √ − 4x + 4 = 1, x = 3/2, y = 3 − x2 = 3/2. So the required tangent lines are
3
y = ±( 3/3)x.
(b) Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve
x2 − 4x + y 2 + 3 = 0. Implicit differentiation applied to the equation of the curve gives
dy/dx = (2 − x)/y. At P the slope of the curve must equal the slope of the line so
(2 − x0 )/y0 = y0 /x0 , or y0 = 2x0 − x2 . But x2 − 4x0 + y0 + 3 = 0 because (x0 , y0 ) is on the
2
0 0
2
curve, and elimination of y0 in the latter two equations gives x2 − 4x0 + (2x0 − x√) + 3 = 0,
2
0
2
0
x0 = 3/2 which when substituted into y0 =√ 0 − x0 yields y0 = 3/4, so y0 = ± 3/2. The
√
2
2x 2 2
√
slopes of the lines are (± 3/2)/(3/2) = ± 3/3 and their equations are y = ( 3/3)x and
√
y = −( 3/3)x.
48. Let P (x0 , y0 ) be a point where a line through the origin is tangent to the curve
2x2 − 4x + y 2 + 1 = 0. Implicit differentiation applied to the equation of the curve gives
dy/dx = (2 − 2x)/y. At P the slope of the curve must equal the slope of the line so
(2 − 2x0 )/y0 = y0 /x0 , or y0 = 2x0 (1 − x0 ). But 2x2 − 4x0 + y0 + 1 = 0 because (x0 , y0 ) is on the
2
0
2
curve, and elimination of y0 in the latter two equations gives 2x0 = 4x0 − 1, x0 = 1/2 which when
2
√
substituted into y0 = 2x0 (1 − x0 ) yields y0 = 1/2, √ y0 = ± 2/2. The slopes of the lines are
√
2
√
2
so √
(± 2/2)/(1/2) = ± 2 and their equations are y = 2x and y = − 2x.
49. The linear equation axr−1 x + by0 y = c is the equation of a line . Implicit differentiation of the
0
r−1
dy dy axr−1
equation of the curve yields raxr−1 + rby r−1 = 0, = − r−1 . At the point (x0 , y0 ) the slope
dx dx by
axr−1
of the line must be − r−1 , which is the slope of . Moreover, the equation of is satisfied by
0
by0
the point (x0 , y0 ), so this point lies on . By the point-slope formula, must be the line tangent
to the curve at (x0 , y0 ).
dy
50. Implicit differentiation of the equation of the curve yields rxr−1 + ry r−1 = 0. At the point (1, 1)
dy dy dx
this becomes r + r = 0, = −1.
dx dx
dy dy dt
51. By the chain rule, = . Use implicit differentiation on 2y 3 t + t3 y = 1 to get
dx dt dx
dy 2y 3 + 3t2 y dt 1 dy 2y 3 + 3t2 y
=− , but = so =− .
dt 6ty 2 + t3 dx cos t dx (6ty 2 + t3 ) cos t
4 1/3 4
52. (a) f (x) = x , f (x) = x−2/3
3 9
7 4/3 28 1/3 28 −2/3
(b) f (x) = x , f (x) = x , f (x) = x
3 9 27
(c) generalize parts (a) and (b) with k = (n − 1) + 1/3 = n − 2/3
7. January 27, 2005 11:44 L24-ch04 Sheet number 7 Page number 133 black
Exercise Set 4.2 133
53. y = rxr−1 , y = r(r − 1)xr−2 so 3x2 r(r − 1)xr−2 + 4x rxr−1 − 2xr = 0,
3r(r − 1)xr + 4rxr − 2xr = 0, (3r2 + r − 2)xr = 0,
3r2 + r − 2 = 0, (3r − 2)(r + 1) = 0; r = −1, 2/3
54. y = rxr−1 , y = r(r − 1)xr−2 so 16x2 r(r − 1)xr−2 + 24x rxr−1 + xr = 0,
16r(r − 1)xr + 24rxr + xr = 0, (16r2 + 8r + 1)xr = 0,
16r2 + 8r + 1 = 0, (4r + 1)2 = 0; r = −1/4
55. We shall find when the curves intersect and check that the slopes are negative reciprocals. For the
intersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x − k)2 + y 2 = k 2 to obtain
1 y−c x−k
cy = kx = (x2 + y 2 ). Thus x2 + y 2 = cy + kx, or y 2 − cy = −x2 + kx, and =− .
2 x y
dy x dy x−k
Differentiating the two families yields (black) =− , and (gray) =− . But it was
dx y−c dx y
proven that these quantities are negative reciprocals of each other.
dy dy
56. Differentiating, we get the equations (black) x + y = 0 and (gray) 2x − 2y = 0. The first
dx dx
y x
says the (black) slope is = − and the second says the (gray) slope is , and these are negative
x y
reciprocals of each other.
EXERCISE SET 4.2
1 1 1 1 1
1. (5) = 2. =
5x x x/3 3 x
1 1 1 1
3. 4. √ √ = √ √
1+x 2+ x 2 x 2 x(2 + x)
1 2x 3x2 − 14x
5. 2−1
(2x) = 2 6.
x x −1 x3 − 7x2 − 3
1 (1 + x2 )(1) − x(2x) 1 − x2
7. 2) 2 )2
=
x/(1 + x (1 + x x(1 + x2 )
1 1−x+1+x 2 d d 2
8. = 9. (2 ln x) = 2 ln x =
(1 + x)/(1 − x) (1 − x)2 1 − x2 dx dx x
1 1 1 1
(ln x)−1/2
2
10. 3 (ln x) 11. = √
x 2 x 2x ln x
1 1 1 1
12. √ √ = 13. ln x + x = 1 + ln x
x2 x 2x x
1 2x2
14. x3 + (3x2 ) ln x = x2 (1 + 3 ln x) 15. 2x log2 (3 − 2x) −
x (3 − 2x) ln 2
3 2 2x − 2
16. log2 (x2 − 2x) + 3x log2 (x2 − 2x)
(x2 − 2x) ln 2
8. January 27, 2005 11:44 L24-ch04 Sheet number 8 Page number 134 black
134 Chapter 4
2x(1 + log x) − x/(ln 10)
17. 18. 1/[x(ln 10)(1 + log x)2 ]
(1 + log x)2
1 1 1 1 1 1
19. = 20.
ln x x x ln x ln(ln(x)) ln x x
1 1
21. (sec2 x) = sec x csc x 22. (− sin x) = − tan x
tan x cos x
1 1 sin(2 ln x) sin(ln x2 )
23. − sin(ln x) 24. 2 sin(ln x) cos(ln x) = =
x x x x
1 cot x
25. 2 (2 sin x cos x) = 2 ln 10
ln 10 sin x
1 2 sin x cos x 2 tan x
26. (−2 sin x cos x) = − =−
(ln 10)(1 − sin x)
2 (ln 10) cos 2x ln 10
d 3 8x 11x2 − 8x + 3
27. 3 ln(x − 1) + 4 ln(x2 + 1) = + 2 =
dx x−1 x +1 (x − 1)(x2 + 1)
d 1 2x3
28. [2 ln cos x + ln(1 + x4 )] = −2 tan x +
dx 2 1 + x4
d 1 3x
29. ln cos x − ln(4 − 3x2 ) = − tan x +
dx 2 4 − 3x2
d 1 1 1 1
30. [ln(x − 1) − ln(x + 1)] = −
dx 2 2 x−1 x+1
1 dy 1 2x
31. ln |y| = ln |x| + ln |1 + x2 |,
3
= x 1 + x2 +
3 dx x 3(1 + x2 )
1 dy 1 x−1 1 1
32. ln |y| = [ln |x − 1| − ln |x + 1|], = 5
−
5 dx 5 x+1 x−1 x+1
1 1
33. ln |y| = ln |x2 − 8| + ln |x3 + 1| − ln |x6 − 7x + 5|
3 2
√
dy (x2 − 8)1/3 x3 + 1 2x 3x2 6x5 − 7
= + − 6
dx x6 − 7x + 5 3(x2 − 8) 2(x3 + 1) x − 7x + 5
1
34. ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| − ln |x|
2
dy sin x cos x tan3 x 3 sec2 x 1
= √ cot x − tan x + −
dx x tan x 2x
√ √
√ 1 dy 10 dy 10
35. f (x) = ex e−1
36. ln y = − 10 ln x, =− , =− √
y dx x dx x1+ 10
ln e 1 d 1
37. (a) logx e = = , [logx e] = −
ln x ln x dx x(ln x)2
ln 2 d ln 2
(b) logx 2 = , [logx 2] = −
ln x dx x(ln x)2
9. January 27, 2005 11:44 L24-ch04 Sheet number 9 Page number 135 black
Exercise Set 4.2 135
ln b ln e 1
38. (a) From loga b = for a, b > 0 it follows that log(1/x) e = =− , hence
ln a ln(1/x) ln x
d 1
log(1/x) e =
dx x(ln x)2
ln e 1 d 1 1 1
(b) log(ln x) e = = , so log(ln x) e = − =−
ln(ln x) ln(ln x) dx (ln(ln x))2 x ln x x(ln x)(ln(ln x))2
1
39. f (x0 ) = f (e−1 ) = −1, f (x) = , f (x0 ) = e, y − (−1) = e(x − 1/e) = ex − 1, y = ex − 2
x
dy 1
40. y0 = log 10 = 1, y = log x = (log e) ln x, = log e ,
dx x=10 10
log e log e
y−1= (x − 10), y = x + 1 − log e
10 10
1 1 1
41. f (x0 ) = f (−e) = 1, f (x) =− , 42. y − ln 2 = − (x + 2), y = − x + ln 2 − 1
x=−e e 2 2
1 1
y − 1 = − (x + e), y = − x
e e
43. Let the equation of the tangent line be y = mx and suppose that it meets the curve at (x0 , y0 ).
1 1 1 ln x0 1
Then m = = and y0 = mx0 = ln x0 . So m = = and ln x0 = 1, x0 = e, m =
x x=x0 x0 x0 x0 e
1
and the equation of the tangent line is y = x.
e
44. Let y = mx + b be a line tangent to the curve at (x0 , y0 ). Then b is the y-intercept and the
1
slope of the tangent line is m = . Moreover, at the point of tangency, mx0 + b = ln x0 or
x0
1
x0 + b = ln x0 , b = ln x0 − 1, as required.
x0
y
45. The area of the triangle P QR, given by |P Q||QR|/2 is
required. |P Q| = w, and, by Exercise 44, |QR| = 1, so 1
P (w, ln w)
area = w/2. Q x
w 2
R
–2
46. Since y = 2 ln x, let y = 2z; then z = ln x and we apply the result of Exercise 45 to find that the
area is, in the x-z plane, w/2. In the x-y plane, since y = 2z, the vertical dimension gets doubled,
so the area is w.
dy 1 dy 1
47. If x = 0 then y = ln e = 1, and = . But ey = x + e, so = y = e−y .
dx x+e dx e
dy 1
. But ey = e− ln(e −x) = (e2 − x)−1 , so
2
48. When x = 0, y = − ln(e2 ) = −2. Next, = 2
dx e −x
dy
= ey .
dx
10. January 27, 2005 11:44 L24-ch04 Sheet number 10 Page number 136 black
136 Chapter 4
dy 1
49. Let y = ln(x + a). Following Exercise 47 we get = = e−y , and when x = 0, y = ln(a) = 0
dx x+a
if a = 1, so let a = 1, then y = ln(x + 1).
dy 1 1 dy
50. Let y = − ln(a − x), then = . But ey = , so = ey .
dx a−x a−x dx
If x = 0 then y = − ln(a) = − ln 2 provided a = 2, so y = − ln(2 − x).
ln(e2 + ∆x) − 2 d 1
51. (a) f (x) = ln x; f (e2 ) = lim = (ln x) = = e−2
∆x→0 ∆x dx x=e2 x x=e2
ln(1 + h) − ln 1 ln(1 + h) 1
(b) f (w) = ln w; f (1) = lim = lim = =1
h→0 h h→0 h w w=1
f (x) − f (0)
52. (a) Let f (x) = ln(cos x), then f (0) = ln(cos 0) = ln 1 = 0, so f (0) = lim =
x→0 x
ln(cos x)
lim , and f (0) = − tan 0 = 0.
x→0 x √
√
2 f (1 + h) − f (1) (1 + h) 2
−1
(b) Let f (x) = x , then f (1) = 1, so f (1) = lim = lim , and
√ √ √ h→0 h h→0 h
f (x) = 2x 2−1 , f (1) = 2.
d logb (x + h) − logb (x)
53. [logb x] = lim
dx h→0 h
1 x+h
= lim logb Theorem 1.6.2(b)
h→0 h x
1 h
= lim logb 1 +
h→0 h x
1
= lim logb (1 + v) Let v = h/x and note that v → 0 as h → 0
v→0 vx
1 1
= lim logb (1 + v) h and v are variable, whereas x is constant
x v→0 v
1
= lim log (1 + v)1/v Theorem 1.6.2.(c)
x v→0 b
1
= logb lim (1 + v)1/v Theorem 2.5.5
x v→0
1
= logb e Formula 7 of Section 7.1
x
EXERCISE SET 4.3
1. (a) f (x) = 5x4 + 3x2 + 1 ≥ 1 so f is one-to-one on −∞ < x < +∞.
d −1 1 1 1
(b) f (1) = 3 so 1 = f −1 (3); f (x) = −1 (x))
, (f −1 ) (3) = =
dx f (f f (1) 9
2. (a) f (x) = 3x2 + 2ex ; for −1 < x < 1, f (x) ≥ 2e−1 = 2/e, and for |x| > 1, f (x) ≥ 3x2 ≥ 3, so
f is increasing and one-to-one
d −1 1 1 1
(b) f (0) = 2 so 0 = f −1 (2); f (x) = , (f −1 ) (2) = =
dx f (f −1 (x)) f (0) 2
11. January 27, 2005 11:44 L24-ch04 Sheet number 11 Page number 137 black
Exercise Set 4.3 137
2 d −1 2
3. f −1 (x) = − 3, so directly f (x) = − 2 . Using Formula (1),
x dx x
−2 1
f (x) = , so = −(1/2)(f −1 (x) + 3)2 ,
(x + 3)2 f (f −1 (x))
2
d −1 2 2
f (x) = −(1/2) =−
dx x x2
ex − 1 d −1 ex 2
4. f −1 (x) = , so directly, f (x) = . Next, f (x) = , and using Formula (1),
2 dx 2 2x + 1
d −1 2f −1 (x) + 1 ex
f (x) = =
dx 2 2
5. (a) f (x) = 2x + 8; f < 0 on (−∞, −4) and f > 0 on (−4, +∞); not enough information. By
inspection, f (1) = 10 = f (−9), so not one-to-one
(b) f (x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f (x) is positive for all x, so f is one-to-one
(c) f (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one
x
(d) f (x) = −(ln 2) 1 < 0 because ln 2 > 0, so f is one-to-one for all x.
2
6. (a) f (x) = 3x2 + 6x = x(3x + 6) changes sign at x = −2, 0, so √ enough information; by
not √
observation (of the graph, and using some guesswork), f (−1 + 3) = −6 = f (−1 − 3), so
f is not one-to-one.
(b) f (x) = 5x4 + 24x2 + 2 ≥ 2 > 0; f is positive for all x, so f is one-to-one
1
(c) f (x) = ; f is one-to-one because:
(x + 1)2
if x1 < x2 < −1 then f > 0 on [x1 , x2 ], so f (x1 ) = f (x2 )
if −1 < x1 < x2 then f > 0 on [x1 , x2 ], so f (x1 ) = f (x2 )
if x1 < −1 < x2 then f (x1 ) > 1 > f (x2 ) since f (x) > 1 on (−∞, −1) and f (x) < 1 on
(−1, +∞)
d 1
(d) Note that f (x) is only defined for x > 0. logb x = , which is always negative
dx x ln b
(0 < b < 1), so f is one-to-one.
dx dy 1
7. y = f −1 (x), x = f (y) = 5y 3 + y − 7, = 15y 2 + 1, = ;
dy dx 15y 2 + 1
dy dy dy 1
check: 1 = 15y 2 + , = 2+1
dx dx dx 15y
dx dy
8. y = f −1 (x), x = f (y) = 1/y 2 , = −2y −3 , = −y 3 /2;
dy dx
dy dy
check: 1 = −2y −3 , = −y 3 /2
dx dx
dx dy 1
9. y = f −1 (x), x = f (y) = 2y 5 + y 3 + 1, = 10y 4 + 3y 2 , = 4 + 3y 2
;
dy dx 10y
dy dy dy 1
check: 1 = 10y 4 + 3y 2 , =
dx dx dx 10y 4 + 3y 2
dx dy 1
10. y = f −1 (x), x = f (y) = 5y − sin 2y, = 5 − 2 cos 2y, = ;
dy dx 5 − 2 cos 2y
dy dy 1
check: 1 = (5 − 2 cos 2y) , =
dx dx 5 − 2 cos 2y
12. January 27, 2005 11:44 L24-ch04 Sheet number 12 Page number 138 black
138 Chapter 4
12. −10xe−5x
2
11. 7e7x
1 1/x
13. x3 ex + 3x2 ex = x2 ex (x + 3) 14. − e
x2
dy (ex + e−x )(ex + e−x ) − (ex − e−x )(ex − e−x )
15. =
dx (ex + e−x )2
(e2x + 2 + e−2x ) − (e2x − 2 + e−2x )
= = 4/(ex + e−x )2
(ex + e−x )2
16. ex cos(ex )
dy (ln x)ex − ex (1/x) ex (x ln x − 1)
17. (x sec2 x + tan x)ex tan x 18. = =
dx (ln x)2 x(ln x)2
15 2
x (1 + 5x3 )−1/2 exp( 1 + 5x3 )
3x
19. (1 − 3e3x )e(x−e )
20.
2
(x − 1)e−x x−1 1
21. = x 22. [− sin(ex )]ex = −ex tan(ex )
1 − xe−x e −x cos(ex )
1
23. f (x) = 2x ln 2; y = 2x , ln y = x ln 2, y = ln 2, y = y ln 2 = 2x ln 2
y
1
24. f (x) = −3−x ln 3; y = 3−x , ln y = −x ln 3, y = − ln 3, y = −y ln 3 = −3−x ln 3
y
25. f (x) = π sin x (ln π) cos x;
1
y = π sin x , ln y = (sin x) ln π, y = (ln π) cos x, y = π sin x (ln π) cos x
y
26. f (x) = π x tan x (ln π)(x sec2 x + tan x);
1
y = π x tan x , ln y = (x tan x) ln π, y = (ln π)(x sec2 x + tan x)
y
y = π x tan x (ln π)(x sec2 x + tan x)
1 dy 3x2 − 2 1
27. ln y = (ln x) ln(x3 − 2x), = 3 ln x + ln(x3 − 2x),
y dx x − 2x x
dy 3x2 − 2 1
= (x3 − 2x)ln x 3 ln x + ln(x3 − 2x)
dx x − 2x x
1 dy sin x dy sin x
28. ln y = (sin x) ln x, = + (cos x) ln x, = xsin x + (cos x) ln x
y dx x dx x
1 dy 1
29. ln y = (tan x) ln(ln x), = tan x + (sec2 x) ln(ln x),
y dx x ln x
dy tan x
= (ln x)tan x + (sec2 x) ln(ln x)
dx x ln x
13. January 27, 2005 11:44 L24-ch04 Sheet number 13 Page number 139 black
Exercise Set 4.3 139
1 dy 2x 1
30. ln y = (ln x) ln(x2 + 3), = 2 ln x + ln(x2 + 3),
y dx x +3 x
dy 2x 1
= (x2 + 3)ln x 2 ln x + ln(x2 + 3)
dx x +3 x
31. f (x) = exe−1
32. (a) because xx is not of the form ax where a is constant
1
(b) y = xx , ln y = x ln x, y = 1 + ln x, y = xx (1 + ln x)
y
3 3 1/2 1
33. =√ 34. − =−
1− (3x)2 1 − 9x2 1− x+1 2 4 − (x + 1)2
2
1 1 sin x sin x 1, sin x > 0
35. (−1/x2 ) = − √ 36. √ = =
1 − 1/x2 |x| x2 − 1 1 − cos2 x | sin x| −1, sin x < 0
3x2 3x2 5x4 5
37. = 38. = √
1 + (x3 )2 1 + x6 |x5 | (x5 )2 − 1 |x| x10 − 1
39. y = 1/ tan x = cot x, dy/dx = − csc2 x
1
40. y = (tan−1 x)−1 , dy/dx = −(tan−1 x)−2
1 + x2
ex 1
41. √ + ex sec−1 x 42. − √
|x| x2 − 1 (cos−1 x) 1 − x2
3x2 (sin−1 x)2
43. 0 44. √ + 2x(sin−1 x)3
1 − x2
√
45. 0 46. −1/ e2x − 1
1 1 −1/2 1 1
47. − x =− √ 48. − √
1+x 2 2(1 + x) x −1
2 cot x(1 + x2 )
49. (a) Let x = f (y) = cot y, 0 < y < π, −∞ < x < +∞. Then f is differentiable and one-to-one
and f (f −1 (x)) = − csc2 (cot−1 x) = −x2 − 1 = 0, and
d 1 1
[cot−1 x] = lim = − lim 2 = −1.
dx x=0
x→0 f (f −1 (x)) x→0 x + 1
(b) If x = 0 then, from Exercise 50(a) of Section 1.5,
d d 1 1 1 1
cot−1 x = tan−1 = − 2 =− 2 . For x = 0, Part (a) shows the same;
dx dx x x 1 + (1/x)2 x +1
d 1
thus for −∞ < x < +∞, [cot−1 x] = − 2 .
dx x +1
d 1 du
(c) For −∞ < u < +∞, by the chain rule it follows that [cot−1 u] = − 2 .
dx u + 1 dx
14. January 27, 2005 11:44 L24-ch04 Sheet number 14 Page number 140 black
140 Chapter 4
d d 1 1 1 −1
50. (a) By the chain rule, [csc−1 x] = sin−1 = − 2 = √
dx dx x x 1− (1/x)2 |x| x2 − 1
d du d −1 du
(b) By the chain rule, [csc−1 u] = [csc−1 u] = √
dx dx du |u| u2 − 1 dx
x (3x2 + tan−1 y)(1 + y 2 )
51. x3 + x tan−1 y = ey , 3x2 + y + tan−1 y = ey y , y =
1+y 2 (1 + y 2 )ey − x
1 1
52. sin−1 (xy) = cos−1 (x − y), (xy + y) = − (1 − y ),
1− x2 y 2 1 − (x − y)2
y 1 − (x − y)2 + 1 − x2 y 2
y =
1 − x2 y 2 − x 1 − (x − y)2
53. (a) f (x) = x3 − 3x2 + 2x = x(x − 1)(x − 2) so f (0) = f (1) = f (2) = 0 thus f is not one-to-one.
√
6 ± 36 − 24 √
(b) f (x) = 3x − 6x + 2, f (x) = 0 when x =
2
= 1 ± 3/3. f (x) > 0 (f is
√ 6 √ √
increasing) if x < 1 − 3/3, f (x) < 0 (f is decreasing) if 1 − 3/3 < x < 1 + 3/3, so f (x)
√ √ √
takes on values less than f (1 − 3/3) on both sides of 1 − 3/3 thus 1 − 3/3 is the largest
value of k.
54. (a) f (x) = x3 (x − 2) so f (0) = f (2) = 0 thus f is not one to one.
(b) f (x) = 4x3 − 6x2 = 4x2 (x − 3/2), f (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2]
and increasing on [3/2, +∞) so 3/2 is the smallest value of k.
55. (a) f (x) = 4x3 + 3x2 = (4x + 3)x2 = 0 only at x = 0. But on [0, 2], f has no sign change, so f
is one-to-one.
(b) F (x) = 2f (2g(x))g (x) so F (3) = 2f (2g(3))g (3). By inspection f (1) = 3, so
g(3) = f −1 (3) = 1 and g (3) = (f −1 ) (3) = 1/f (f −1 (3)) = 1/f (1) = 1/7 because
f (x) = 4x3 + 3x2 . Thus F (3) = 2f (2)(1/7) = 2(44)(1/7) = 88/7.
F (3) = f (2g(3)) = f (2·1) = f (2) = 24, so the line tangent to F (x) at (3, 25) has the equation
y − 25 = (88/7)(x − 3), y = (88/7)x − 89/7.
2 1
56. (a) f (x) = −e4−x 2+ < 0 for all x > 0, so f is one-to-one.
x2
(b) By inspection, f (2) = 1/2, so 2 = f −1 (1/2) = g(1/2). By inspection,
1 9
f (2) = − 2 + = − , and
4 4
d
F (1/2) = f ([g(x)]2 ) [g(x)2 ] = f ([g(x)]2 )2g(x)g (x)
dx x=1/2 x=1/2
−12 1
1 f (4) e (2 + 16 ) 33 11
= f (22 )2 · 2 =4 =4 = = 12
f (g(x)) x=1/2 f (2) (2 + 1 )
4
9e12 3e
57. (a) f (x) = kekx , f (x) = k 2 ekx , f (x) = k 3 ekx , . . . , f (n) (x) = k n ekx
(b) g (x) = −ke−kx , g (x) = k 2 e−kx , g (x) = −k 3 e−kx , . . . , g (n) (x) = (−1)n k n e−kx
dy
58. = e−λt (ωA cos ωt − ωB sin ωt) + (−λ)e−λt (A sin ωt + B cos ωt)
dt
= e−λt [(ωA − λB) cos ωt − (ωB + λA) sin ωt]
15. January 27, 2005 11:44 L24-ch04 Sheet number 15 Page number 141 black
Exercise Set 4.3 141
2 2
1 1 x−µ d 1 x−µ
59. f (x) = √ exp − −
2πσ 2 σ dx 2 σ
2
1 1 x−µ x−µ 1
=√ exp − −
2πσ 2 σ σ σ
2
1 1 x−µ
= −√ (x − µ) exp −
2πσ 3 2 σ
60. y = Aekt , dy/dt = kAekt = k(Aekt ) = ky
61. y = Ae2x + Be−4x , y = 2Ae2x − 4Be−4x , y = 4Ae2x + 16Be−4x so
y + 2y − 8y = (4Ae2x + 16Be−4x ) + 2(2Ae2x − 4Be−4x ) − 8(Ae2x + Be−4x ) = 0
62. (a) y = −xe−x + e−x = e−x (1 − x), xy = xe−x (1 − x) = y(1 − x)
(b) y = −x2 e−x + e−x = e−x (1 − x2 ), xy = xe−x
2 2 2 2
/2 /2 /2 /2
(1 − x2 ) = y(1 − x2 )
dy
63. = 100(−0.2)e−0.2x = −20y, k = −0.2
dx
64. ln y = (5x + 1) ln 3 − (x/2) ln 4, so
y /y = 5 ln 3 − (1/2) ln 4 = 5 ln 3 − ln 2, and
y = (5 ln 3 − ln 2)y
y 7e−t 7e−t + 5 − 5 1
65. ln y = ln 60 − ln(5 + 7e−t ), = = = 1 − y, so
y 5 + 7e−t 5 + 7e−t 12
dy y
=r 1− y, with r = 1, K = 12.
dt K
66. (a) 12
0 9
0
60 60 60
(b) P tends to 12 as t gets large; lim P (t) = lim = = = 12
t→+∞ t→+∞ 5 + 7e−t 5 + 7 lim e−t 5
t→+∞
(c) the rate of population growth tends to zero
3.2
0 9
0
10h − 1 d x d x ln 10
67. lim = 10 = e = ln 10
h→0 h dx x=0 dx x=0
16. January 27, 2005 11:44 L24-ch04 Sheet number 16 Page number 142 black
142 Chapter 4
tan−1 (1 + h) − π/4 d 1 1
68. lim = tan−1 x = =
h→0 h dx x=1 1 + x2 x=1 2
√
9[sin−1 ( 23 + ∆x)]2 − π 2 d 3
69. lim = (3 sin−1 x)2 √
= 2(3 sin−1 x) √ √
∆x→0 ∆x dx x= 3/2 1 − x2 x= 3/2
π 3
= 2(3 ) = 12π
3 1 − (3/4)
(2 + ∆x)(2+∆x) − 4 d x d x ln x
70. lim = x = e
∆x→0 ∆x dx x=2 dx x=2
= (1 + ln x)ex ln x = (1 + ln 2)22 = 4(1 + ln 2)
x=2
√
3 sec−1 w − π d 3 3
71. lim = 3 sec−1 x = √ =
w→2 w−2 dx x=2 |2| 2 2−1 2
4(tan−1 w)w − π d d x ln tan−1 x
72. lim = 4(tan−1 x)x = 4e
w→1 w−1 dx x=1 dx x=1
2
1/(1 + x ) 14
= 4(tan−1 x)x ln tan−1 x + x = π ln(π/4) + == 2 + π ln(π/4)
tan−1 x x=1 2π
EXERCISE SET 4.4
x2 − 4 (x − 2)(x + 2) x+2 2
1. (a) lim = lim = lim =
x→2 x2 + 2x − 8 x→2 (x + 4)(x − 2) x→2 x + 4 3
5
2x − 5 2 − lim 2
x→+∞ x
(b) lim = =
x→+∞ 3x + 7 7 3
3 + lim
x→+∞ x
sin x cos x sin x
2. (a) = sin x = cos x so lim = lim cos x = 1
tan x sin x x→0 tan x x→0
x2 − 1 (x − 1)(x + 1) x+1 x2 − 1 2
(b) 3−1
= = 2 so lim 3 =
x (x − 1)(x 2 + x + 1) x +x+1 x→1 x − 1 3
π π
3. Tf (x) = −2(x + 1), Tg (x) = −3(x + 1), 4. Tf (x) = − x − , Tg (x) = − x −
2 2
limit = 2/3 limit = 1
x
e 1
5. lim =1 6. lim = 1/5
x→0 cos x x→3 6x − 13
sec2 θ tet + et
7. lim =1 8. lim = −1
θ→0 1 t→0 −et
cos x cos x
9. lim+ = −1 10. lim = +∞
x→π 1 x→0+ 2x
1/x 3e3x 9e3x
11. lim =0 12. lim = lim = +∞
x→+∞ 1 x→+∞ 2x x→+∞ 2
17. January 27, 2005 11:44 L24-ch04 Sheet number 17 Page number 143 black
Exercise Set 4.4 143
− csc2 x −x −1
13. lim+ = lim 2 = lim+ = −∞
x→0 1/x x→0 sin x
+ x→0 2 sin x cos x
−1/x x
14. lim = lim 1/x = 0
x→0+ (−1/x2 )e1/x x→0+ e
100x99 (100)(99)x98 (100)(99)(98) · · · (1)
15. lim x
= lim x
= · · · = lim =0
x→+∞ e x→+∞ e x→+∞ ex
√
cos x/ sin x 2/ 1 − 4x2
16. lim = lim+ cos2 x = 1 17. lim =2
x→0+ sec2 x/ tan x x→0 x→0 1
1
1− 1 1 x 1
18. lim 1 + x2 = lim = 19. lim xe−x = lim = lim x = 0
x→0 3x 2 x→0 3(1 + x2 ) 3 x→+∞ x→+∞ ex x→+∞ e
x−π 1
20. lim (x − π) tan(x/2) = lim = lim = −2
x→π x→π cot(x/2) x→π −(1/2) csc2 (x/2)
sin(π/x) (−π/x2 ) cos(π/x)
21. lim x sin(π/x) = lim = lim = lim π cos(π/x) = π
x→+∞ x→+∞ 1/x x→+∞ −1/x2 x→+∞
ln x 1/x − sin2 x −2 sin x cos x
22. lim tan x ln x = lim = lim+ = lim+ = lim+ =0
x→0+ x→0+ cot x x→0 − csc2 x x→0 x x→0 1
cos 5x −5 sin 5x −5(+1) 5
23. lim sec 3x cos 5x = lim = lim = =−
x→(π/2)− x→(π/2)− cos 3x x→(π/2)− −3 sin 3x (−3)(−1) 3
x−π 1
24. lim (x − π) cot x = lim = lim =1
x→π x→π tan x x→π sec2 x
ln(1 − 3/x) −3
25. y = (1 − 3/x)x , lim ln y = lim = lim = −3, lim y = e−3
x→+∞ x→+∞ 1/x x→+∞ 1 − 3/x x→+∞
3 ln(1 + 2x) 6
26. y = (1 + 2x)−3/x , lim ln y = lim − = lim − = −6, lim y = e−6
x→0 x→0 x x→0 1 + 2x x→0
ln(ex + x) ex + 1
27. y = (ex + x)1/x , lim ln y = lim = lim x = 2, lim y = e2
x→0 x→0 x x→0 e + x x→0
b ln(1 + a/x) ab
28. y = (1 + a/x)bx , lim ln y = lim = lim = ab, lim y = eab
x→+∞ x→+∞ 1/x x→+∞ 1 + a/x x→+∞
ln(2 − x) 2 sin2 (πx/2)
29. y = (2 − x)tan(πx/2) , lim ln y = lim = lim = 2/π, lim y = e2/π
x→1 x→1 cot(πx/2) x→1 π(2 − x) x→1
2 ln cos(2/x) (−2/x2 )(− tan(2/x))
30. y = [cos(2/x)]x , lim ln y = lim = lim
x→+∞ x→+∞ 1/x2 x→+∞ −2/x3
− tan(2/x) (2/x2 ) sec2 (2/x)
= lim = lim = −2, lim y = e−2
x→+∞ 1/x x→+∞ −1/x2 x→+∞
1 1 x − sin x 1 − cos x sin x
31. lim − = lim = lim = lim =0
x→0 sin x x x→0 x sin x x→0 x cos x + sin x x→0 2 cos x − x sin x
18. January 27, 2005 11:44 L24-ch04 Sheet number 18 Page number 144 black
144 Chapter 4
1 − cos 3x 3 sin 3x 9 9
32. lim 2
= lim = lim cos 3x =
x→0 x x→0 2x x→0 2 2
(x2 + x) − x2 x 1
33. lim √ = lim √ = lim = 1/2
x→+∞ x2+x+x x→+∞ x2+x+x x→+∞ 1 + 1/x + 1
ex − 1 − x ex − 1 ex
34. lim = lim x = lim x = 1/2
x→0 xex − x x→0 xe + ex − 1 x→0 xe + 2ex
ex
35. lim [x − ln(x2 + 1)] = lim [ln ex − ln(x2 + 1)] = lim ln ,
x→+∞ x→+∞ x→+∞ x2 + 1
x x x
e e e
lim = lim = lim = +∞ so lim [x − ln(x2 + 1)] = +∞
x→+∞ x2 + 1 x→+∞ 2x x→+∞ 2 x→+∞
x 1
36. lim ln = lim ln = ln(1) = 0
x→+∞ 1 + x x→+∞ 1/x + 1
ln x 1/x 1
38. (a) lim = lim = lim =0
xn
x→+∞ x→+∞ nxn−1 x→+∞ nxn
xn nxn−1
(b) lim = lim = lim nxn = +∞
x→+∞ ln x x→+∞ 1/x x→+∞
3x2 − 2x + 1 0
39. (a) L’Hˆpital’s Rule does not apply to the problem lim
o because it is not a form.
x→1 3x2 − 2x 0
3x2 − 2x + 1
(b) lim =2
x→1 3x2 − 2x
e3x −12x+12
2
e0
40. L’Hˆpital’s Rule does not apply to the problem
o , which is of the form , and from
x4 − 16 0
which it follows that lim− and lim+ exist, with values −∞ if x approaches 2 from the left and
x→2 x→2
+∞ if from the right. The general limit lim does not exist.
x→2
1/(x ln x) 2
41. lim √ = lim √ =0 0.15
x→+∞ 1/(2 x) x→+∞ x ln x
100 10000
0
ln x
42. y = xx , lim ln y = lim = lim −x = 0, lim y = 1 1
x→0+ x→0+ 1/x x→0+ x→0+
0 0.5
0
19. January 27, 2005 11:44 L24-ch04 Sheet number 19 Page number 145 black
Exercise Set 4.4 145
43. y = (sin x)3/ ln x , 25
3 ln sin x x
lim ln y = lim = lim (3 cos x) = 3,
x→0+ x→0+ ln x x→0+ sin x
lim y = e3
x→0+
0 0.5
19
4 sec2 x 4 4.1
44. lim − = lim =4
x→π/2 sec x tan x x→π/2− sin x
1.4 1.6
3.3
1 e−x ln x − 1
45. ln x − ex = ln x − = ; 0
e−x e−x 0 3
ln x 1/x
lim e−x ln x = lim = lim = 0 by L’Hˆpital’s Rule,
o
x→+∞ x→+∞ ex x→+∞ ex
e−x ln x − 1
so lim [ln x − ex ] = lim = −∞
x→+∞ x→+∞ e−x
–16
ex
46. lim [ln ex − ln(1 + 2ex )] = lim ln –0.6
x→+∞ x→+∞ 1 + 2ex 0 12
1 1
= lim ln = ln ;
x→+∞ e−x +2 2
horizontal asymptote y = − ln 2
–1.2
1.02
47. y = (ln x)1/x ,
ln(ln x) 1
lim ln y = lim = lim = 0;
x→+∞ x→+∞ x x→+∞ x ln x
lim y = 1, y = 1 is the horizontal asymptote
x→+∞
100 10000
1
20. January 27, 2005 11:44 L24-ch04 Sheet number 20 Page number 146 black
146 Chapter 4
x+1
x+1
x ln
48. y = , lim ln y = lim x+2 1
x+2 x→+∞ x→+∞ 1/x
−x2
= lim = −1;
x→+∞ (x + 1)(x + 2)
lim y = e−1 is the horizontal asymptote
x→+∞
0 50
0
49. (a) 0 (b) +∞ (c) 0 (d) −∞ (e) +∞ (f ) −∞
(ln a) ln x (ln a)/x
50. (a) Type 00 ; y = x(ln a)/(1+ln x) ; lim ln y = lim = lim+ = lim+ ln a = ln a,
x→0+ x→0+ 1 + ln x x→0 1/x x→0
lim y = eln a = a
x→0+
(b) Type ∞0 ; same calculation as Part (a) with x → +∞
(ln a) ln(x + 1) ln a
(c) Type 1∞ ; y = (x + 1)(ln a)/x , lim ln y = lim = lim = ln a,
x→0 x→0 x x→0 x + 1
lim y = eln a = a
x→0
1 + 2 cos 2x x + sin 2x sin 2x
51. lim does not exist, nor is it ±∞; lim = lim 1+ =1
x→+∞ 1 x→+∞ x x→+∞ x
2 − cos x 2x − sin x 2 − (sin x)/x 2
52. lim does not exist, nor is it ±∞; lim = lim =
x→+∞ 3 + cos x x→+∞ 3x + sin x x→+∞ 3 + (sin x)/x 3
x(2 + sin 2x) 2 + sin 2x
53. lim (2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim = lim ,
x→+∞ x+1
x→+∞ x→+∞ 1 + 1/x
which does not exist because sin 2x oscillates between −1 and 1 as x → +∞
1 1 sin x
54. lim + cos x + does not exist, nor is it ±∞;
x→+∞ x 2 2x
x(2 + sin x) 2 + sin x
lim = lim =0
x→+∞ x2 + 1 x→+∞ x + 1/x
V t −Rt/L
L e Vt
55. lim+ =
R→0 1 L
π/2 − x −1
56. (a) lim (π/2 − x) tan x = lim = lim = lim sin2 x = 1
x→π/2 x→π/2 cot x x→π/2 − csc2 x x→π/2
1 1 sin x cos x − (π/2 − x) sin x
(b) lim − tan x = lim − = lim
x→π/2 π/2 − x x→π/2 π/2 − x cos x x→π/2 (π/2 − x) cos x
−(π/2 − x) cos x
= lim
x→π/2 −(π/2 − x) sin x − cos x
(π/2 − x) sin x + cos x
= lim =0
x→π/2 −(π/2 − x) cos x + 2 sin x
(c) 1/(π/2 − 1.57) ≈ 1255.765534, tan 1.57 ≈ 1255.765592;
1/(π/2 − 1.57) − tan 1.57 ≈ 0.000058